A) 10.30
B) 3.70
C) 11.30
D) None of these
Correct Answer: C
Solution :
[c] \[[{{H}^{+}}]\] in HCl solution (pH = 2) = \[{{10}^{-2}}\] M ; \[[O{{H}^{-}}]\] in KOH solution \[(pOH=14-12=2)={{10}^{-2}}M\] Excess m Mol of \[O{{H}^{-}}\] in 5 ml mixture \[3\times {{10}^{-2}}-2\times {{10}^{-2}}=1.0\times {{10}^{-2}};\] \[[O{{H}^{-}}]\]in mixture\[=\frac{1.0\times {{10}^{-2}}}{5}\] \[=2\times {{10}^{-3}}\,\,M;\] pOH \[=-\log 2\times {{10}^{-3}}=3-\log 2;\] pH\[=14-(3-\log 2)=11.30\]
You need to login to perform this action.
You will be redirected in
3 sec
